# A step in the proof of the Riemann Mapping Theorem

Let $\Omega \subsetneq \Bbb C$ be open and simply connected. Let $\overline{\Bbb C}$ denote the Riemann sphere and assume without loss of generality that $0 \in \overline{\Bbb C} \backslash \Omega$. Is there some result that allows us to conclude that $\exists$ a path $\gamma$ in $\overline{\Bbb C} \backslash \Omega$ between $0$ and $\infty$? I am positive that this is true but I am unaware of what allows us to conclude this. An explanation from the view-point of point-set topology will be particularly appreciated but if such does not exist, any relatively understandable explanation will do. Please recall that you cannot quote the Riemann Mapping Theorem as this is what we are trying to prove.

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Take $A$ to be the closure of the graph of $y=\sin(1/x), x>0$, and $\Omega$ to be the complement of $A$. I'm pretty sure $\Omega$ is a counterexample, but can't yet see a nice way to prove it's simply connected. – Chris Eagle Jul 4 '13 at 11:02
But if you take complement then the open set that you get is that connected? – Abelvikram Jul 4 '13 at 11:12
I think @ChrisEagle 's example works. the complement of the closed topologist's sine curve should be path connected and simply connected. – Dan Rust Jul 4 '13 at 11:13
Any point in the complement of $A$ should be able to be connected to the 'standard' part of the plane by a vertical interval. – Dan Rust Jul 4 '13 at 11:18
Bak-Newman's Complex Analysis book (section 8.1) has a definition of simple-connectedness which is "A region is simply connected if its complement is connected within $\epsilon$ to $\infty$". By this they mean that for each point in the complement and each $\epsilon$ there is a path connecting the point to $\infty$ which stays within $\epsilon$ of the complement. The closure of the topologist's sine curve is then provided as a counterexample to a possible definition requiring the path to strictly stay in the complement. – bryanj Jul 4 '13 at 11:50

Bak-Newman's Complex Analysis book (section 8.1) has a definition of simple-connectedness which is "A region is simply connected if its complement is connected within $\epsilon$ to $\infty$". By this they mean that for each point in the complement and each $\epsilon$ there is a path connecting the point to $\infty$ which stays within $\epsilon$ of the complement. The closure of the topologist's sine curve is then provided as a counterexample to a possible definition requiring the path to strictly stay in the complement.